Ordinary.differential.equations Tenenbaum Pollard 0486649407

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Differential Equations for University Level

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  • ORDINARY DIFFERENTIAL EQUATIONS

    An Elementary Textbook for Students of Mathematics, Engineering, and the Sciences

    Morris Tenenbaum Harry Pollard

  • ORDINARY DIFFERENTIAL EQUATIONS

    An Elementary Textbook for Students of Mathematics, Engineering,

    and the Sciences

    Morris Tenenbamn CorneU University

    Harry Pollard Purdue University

    DOVER PUBLICATIONS, INC., NEW YORK

  • Copyright 1963 by Morris ~nenbaum and Harry Pollard All rights reserved under Pan American and International Copyright Con-

    ventions.

    This Dover edition, first published in 1985, is an unabridged and corrected republication of the work first published by Harper & Row, Publishers, Inc., New York, in 1963.

    Manufactured in the United States of America Dover Publications, Inc., 31 East 2nd Street, Mineola, N.Y. 11501

    Library of Congress Cataloging in Publication Data Tenenbaum, Morris.

    Ordinary differential equations. Reprint. Originally published: New York: Harper & Row, 1963. Bibliography: p. Includes index. 1. Differential equations. I. Pollard, Harry, 1919- II. Title.

    QA372.T4 1985 515.3'5 85-12983 ISBN 0-486-64940-7

  • Contents

    PREFACE FOR THE TEACHER :n'

    PREFACE FOR THE STUDENT XvU

    1. BASIC CONCEPTS 1

    Lesson I. How Differential Equations Originate. I

    Lesson 2. The Meaning of the Terms Set and Function. Im-plicit Functions. Elementary Functions. 5

    A. The Meaning of the Term Set. 5 B. The Meaning of the Term Function of One !?~dependent Variable. 6 C. Function of Two Inde-pendent Variables. 11 D. Implicit Function. 14 E. The Elemen-tary Functions. 17

    Lesson 3. The Differential Equation. 20 A. Definition of an Ordinary Differential Equation. Order of a Differ-ential Equation. BO B. Solution of a Differential Equation. Explicit Solution. B1 C. Implicit Solution of a Differential Equation. B4

    Lesson 4. The General Solution of a Differential Equation. 28 A. Multiplicity of Solutions of a Differential Equation. BB B. Method of Finding a Differential Equation if Its n-Parameter Family of Solu-tions Is Known. S1 C. General Solution. Particular Solution. Initial Conditions. SS

    Lesson 5. Direction Field. 38 A. Construction of a Direction Field. The Isoclines of a Direction Field. SB B. The Ordinary and Singular Points of the First Order Equation (5.11). 41

    2. SPECIAL TYPES OF DIFFERENTIAL EQUATIONS OF THE FIRST ORDER

    Lesson 6. Meaning of the Differential of a Function. Separable Differential Equations.

    A. Differential of a Function of One Independent Variable. 47 B. Dif-ferential of a Function of Two Independent Variables. 50 C. Differ-ential Equations with Separable Variables. 51

    v

    47

  • vi CONTENTS

    Lesson 7. First Order Differential Equation with Homogeneous Coefficients.

    A. Definition of a Homogeneous Function. 57 B. Solution of a Dif-ferential Equation in Which the Coefficients of d:t: and dy Are Each Homogeneous Functions of the Same Order. 58

    57

    Lesson 8. Differential Equations with Linear Coefficients. 62 A. A Review of Some Plane Analytic Geometry. 61J B. Solution of a Differential Equation in Which the Coefficients of d:t: and dy are Linear, Nonhomogeneous, and When Equated to Zero Represent Non-parallel Lines. 63 C. A Second Method of Solving the Differential Equation (8.2) with Nonhomogeneous Coefficients. 66 D. Solution of a Differential Equation in Which the Coefficients of th and dy Define Parallt>l or Coincident Lines. 67

    Lesson 9. Exact Differential Equations. A. Definition of an Exact Differential and of an Exact Differential Equation. 71J B. Necessary and Sufficient Condition for Exactness and Method of Solving an Exact Differential Equation. 73

    Lesson 10. Recognizable Exact Differential Equations. Integrating Factors.

    A. Recognizable Exact Differential Equations. 80 B. Integrating Factors. BIJ C. Finding an Integrating Factor. 84

    Lesson II. The Linear Pifferential Equation of the First Order.

    70

    80

    Bernoulli Equation. 91 A. Definition of a Linear Differential Equation of the First Order. 91 B. Method of Solution of a Linear Differential Equation of the First Order. 91J C. Determination of the Integrating Factor efPth. 94 D. Bernoulli Equation. 95

    Lesson 12. Miscellaneous Methods of Solving a First Order Differential Equation. 99

    A. Equations Permitting a Choice of Method. 99 Substitution and Other Means. 101

    B. Solution by

    3. PROBLEMS LEADING TO DIFFERENTIAL EQUATIONS OF THE FIRST ORDER 107

    Lesson 13. Geometric Problems. 107

    Lesson 14. Trajectories. liS A. Isogonal Trajectories. 115 B. Orthogonal Trajectories. 117 C. Orthogonal Trajectory Formula in Polar Coordinates. 118

  • CONTENTS vii

    Lesson 15. Dilution and Accretion Problems. Interest Prob-lems. Tentperature Problems. Decomposition and Growth Problems. Second Order Processes. 122

    A. Dilution and Accretion Problems. 122 B. Interest Problems. 126 C. Temperature Problems. 129 D. Decomposition and Growth Problems. 131 E. Second Order Processes. 134

    Lesson 16. Motion of a Particle Along a Straight Line-Vertical, Horizontal, Inclined.

    A. Vertical Motion. 139 B. Horizontal Motion. 160 C. Inclined Motion. 164

    Lesson 17. Pursuit Curves. Relative Pursuit Curves. A. Pursuit Curves. 168 B. Relative Pursuit Curve. 177

    Lesson 17M. Miscellaneous Types of Problems Leading to

    138

    168

    Equations of the First Order 183 A. Flow of Water Through an Orifice. 183 B. First Order Linear Electric Circuit. 184 C. Steady State Flow of Heat. 185 D. Pres-sure-Atmospheric and Oceanic. 186 E. RQpe or Chain Around a Cylinder. 188 F. Motion of a Complex System. 189 G. Variable Mass. Rocket Motion. 191 II. Rotation of the Liquid in a Cylin-der. 193

    4. LINEAR DIFFERENTIAL EQUATIONS OF ORDER GREATER THAN ONE

    Lesson 18. Complex Nuntbers and Complex Functions. A. Complex Numbers. 197 B. Algebra of Complex Numbers. 200 C. Exponential, Trigonometric, and Hyperbolic Functions of Complex Numbers. 201

    Lesson 19. Linear Independence of Functions. The Linear

    196

    197

    Differential Equation of Order n. 205 A. Linear Independence of Functions. 205 ential Equation of Order n. 207

    B. The Linear Differ-

    Lesson 20. Solution of the Homogeneous Linear Differential Equation of Order n with Constant Coefficients. 21l

    A. General Form of Its Solutions. 211 B. Roots of the Characteristic Equation (20.14) Real and Distinct. 213 C. Roots of Characteristic Equation (20.14) Real but Some Multiple. 214 D. Some or All Roots of the Characteristic Equation (20.14) Imaginary. 217

  • viii CoNTENTS

    Lesson 21. Solution of the Nonhomogeneous Linear Differential Equation of Order n with Constant Coefficients. 221

    A. Solution by the Method of Undetermined Coefficients. BBl B. So-lution by the Use of Complex Variables. 1830

    Lesson 22. Solution of the Nonhomogeneous Linear Differential Equation by the Method of Variation of Parameters. 233

    A. Introductory Remarks. 18SS B. The Method of Variation of Parameters. :ess

    Lesson 23. Solution of the Linear Differential Equation with Nonconstant Coefficients. Reduction of Order Method. 241

    A. Introductory Remarks. 1841 B. Solution of the Linear Differential Equation with Nonconstsnt Coefficients by t.he Reduction of Order Method. B4B

    5. OPERATORS AND LAPLACE TRANSFORMS 250

    Lesson 24. Differential and Polynomial Operators. 251 A. Definition of an Operator. Linear Property of Polynomial Opera-tors. 151 B. Algebraic Properties of Polynomial Operators. 155 C. Exponential Shift Theorem for Polynomial Operators. B60 D. So-lution of a Linear Differential Equation with Constant Coefficients by Means of Polynomial Operators. 186B

    Lesson 25. Inverse Operators. 268 A. Meaning of an Inverse Operator. 1869 B. Solution of (25.1) by Means of Inverse Operators. B7B

    Lesson 26. Solution of a Linear Differential Equation by Means of the Partial Fraction Expansion of Inverse Operators. 283

    A. Partial Fraction Expansion Theorem. B8S B. First Method of Solving a Linear Equation by Means of the Partial Fraction Expansion of Inverse Operators. 1888 C. A Second Method of Solving a Linear Equation by Means of the Partial Fraction Expansion of Inverse Operators. 190

    Lesson 27. The Laplace Transform. Gamma Function. A. Improper Integral. Definition of a Laplace Transform. 1891 B. Prop-erties of the Laplace Transform. 1895 C. Solution of a Linear Equa-tion with Constant Coefficients by Means of a Laplace Transform. 196 D. Construction of a Table of Laplace Transforms. SOB E. The Gamma Function. S06

    292

  • CONTENTS ix

    6. PROBLEMS LEADING TO LINEAR DIFFERENTIAL EQUATIONS OF ORDER TWO

    Lesson 28. Undamped Motion. A. Free Undamped Motion. (Simple Harmonic Motion.) 313 B. Defi-nitions in Connection with Simple Harmonic Motion. 317 C. Exam-ples of Particles Executing Simple Harmonic Motion. Harmonic Oscil-lators. 323 D. Forced Undamped Motion. 338

    Lesson 29. Damped Motion. A. Free Damped Motion. (Damped Harmonic Motion.) 347 B. Forced Motion with Damping. 359

    Lesson 30. Electric Circuits. Analog Computation. A. Simple Electric Circuit. 369 B. Analog Computation. 375

    Lesson 30M. Miscellaneous Types of Problems Leading to

    313

    313

    347

    369

    Linear Equations of the Second Order 380 A. Problems Involving a Centrifugal Force. 380 B. Rolling Bodies. 381 C. Twisting Bodies. 383 D. Bending of Beams. 383

    7. SYSTEMS OF DIFFERENTIAL EQUATIONS. LINEARIZATION OF FIRST ORDER SYSTEMS 393

    Lesson 31. Solution of a System of Differential Equations. 393 A. Meaning of a Solution of a System of Differential Equations. 393 B. Definition and Solution of a System of First Order Equations. 394 C. Definition and Solution of a System of Linear First Order Equa-tions. 396 D. Solution of a System of Linear Equations with Con-stant Coefficients by the Use of Operators. Nondegenerate Case. 398 E. An Equivalent Triangular System. 405 F. Degenerate Case. f,(D)g,(D) - g,(D)/o(D) - 0. 413 G. Systems of Three Linear Equations. 415 H. Solution of a System of Linear Differential Equa-tions with Constant Coefficients by Means of Laplace Transforms. 418

    Lesson 32. Linearization of First Order'Systems. 424

    8. PROBLEMS GIVING RISE TO SYSTEMS OF EQUATIONS. SPECIAL TYPES OF SECOND ORDER LINEAR AND NON-LINEAR EQUATIONS SOLVABLE BY REDUCING TO SYSTEMS 440

    Lesson 33. Mechanical, Biological, Electrical Problems Giving Rise to Systems of Equations. 440

    A. A Mechanical Problem-Coupled Springs. 440 B. A Biological Problem. 447 C. An Electrical Problem. More Complex Circuits. 451

  • x CoNTENTs

    Lesson 34. Plane Motions Giving Rise to Systems of Equations. 459

    A. Derivation of Velocity and Acceleration Formulas. 1,59 B. The Plane Motion of a Projectile. 463 C. Definition of a Central Force. Properties of the Motion of a Particle Subject to a Central Force. 470 D. Definitions of Fqrce Field, Potential, ConBervative Field. Conser-vation of Energy in a Conservative Field. 473 E. Path of a Particle in Motion Subject to a Central Force Whose Magnitude Is Proportional to Its Distance from a Fixed Point 0. 476 F. Path of a Particle in Motion Subject to a Central Force Whose Magnitude Is Inversely Pro-portional to the Square of Its Distance from a Fixed Point 0. 481 G. Planetary Motion. 491 H. Kepler's (1571-1630) Laws of Plane-tary Motion. Proof of Newton's Inverse Square Law. 49t

    Lesson 35. Special Types of Second Order Linear and Nonlinear Differential Equations Solvable by Reduction to a System of Two First Order Equations. 500

    A. Solution of a Second Order Nonlinear Differential Equation in Which y' and the Independent Variable z Are Absent. 500 B. Solu-tion of a Second Order Nonlinear Differential Equation in Which the Dependent Variable y Is Absent. 50t C. Solution of a Second Order Nonlinear Equation in Which the Independent Variable xIs Absent. 503

    Lesson 36. Problems Giving Rise to Special Types of Second Order Nonlinear Equations. 506

    A. The Suspension Cable. 506 B. A Special Central Force Prob-lem. 5t1 C. A Pursuit froblem Leading to a Second Order Nonlinear Differential Equation. 5t3 D. Geometric Problems. 5t8

    9. SERIES METHODS

    Lesson 37. Power Series Solutions of Linear Differential

    531

    Equations. 531

    A. Review of Taylor Series and Related Matters. 531 B. Solution of Linear Differential Equations by Series Methods. 537

    Lesson 38. Series Solution of y' = f(x,y). 548

    Lesson 39. Series Solution of a Nonlinear Differential Equation of Order Greater Than One and of a System of First Order Differential Equations. 555

    A. Series Solution of a System of First Order Differential Equations. 555 B. Series Solution of a System of Linear First Order Equations. 559 C. Series Solution of a Nonlinear Differential Equation of Order Greater Than One. 56t

  • CONTENTS :d

    Les110n 40. Ordinary Points and Singularities of a Linear Differential Equation. Method of Frobenius. 570

    A. Ordinary Points and Singularities of a Linear Differential Equa-tion. 570 B. Solution of a Homogeneous Linear Differential Equation About a Regular Singularity. Method of Frobenius. 57B

    Les110n 41. The Legendre Differential Equation. Legendre Functions. Legendre Polynomials Plc(z). Properties of Legendre Polynomials Plc(z). 591

    A. The Legendre Differential Equation. 591 B. Comments on the Solution (41.18) of the Legendre Equation (41.1). Legendre Functions. Legendre Polynomials P~(z). 593 C. Properties of Legendre Poly-nomials P~(z). 598

    Lesson 42. The Bessel Differential Equation. Bessel Function of the First Kind Jlc(z). Differential Equations Leading to a Bessel Equation. Properties of Jlc(z). 609

    A. The Bessel Differential Equation. 609 B. Bessel Functions of the First Kind J~(z). 611 C. Differential Equations Which Lead to a Bessel Equation. 615 D. Properties of Bessel Functions of the First Kind h(z). 619

    Les110n 43. The Laguerre Differential Equation. Laguerre Polynomials L~(z). Properties of Llc(z). 624

    A. The Laguerre Differential Equation and Its Solution. 6B4 B. The Laguerre Polynomial L~(z). 6t5 C. Some Properties of Laguerre Polynomials L~(z). 6t7

    10. NUMERICAL METHODS 631

    LesBOn 44. Starting Method. Polygonal Approximation. 632

    Les110n 45. An Improvement of the Polygonal Starting Method. 641

    Les110n 46. Starting Method-Taylor Series. 645 A. Numerical Solution of y' =- /(z,y) by Direct Substitution in a Taylor Series. 64-6 B. Numerical Solution of y' = /(z,y) by the "Creeping Up" Process. 64-6

    Les110n 47. Starting Method-Runge-Kutta Formulas.

    Lesson 48. Finite Differences. Interpolation. 659 A. Finite Differences. 659 B. Polynomial Interpolation. 661

  • Di CoNTENTS

    Lenon 49. Newton's Interpolation Formulas. A. Newton's (Forward) Interpolation Formula. 663 B. Newton's (Backward) Interpolation Formula. 668 C. The Error in Polyno-mial Interpolation. 670

    Lesson 50. Approximation Formulas Including Simpson's and

    663

    Weddle's Rule. 672

    Lesson 51. Milne's Method of Finding an Approximate Numerical Solution of y' = f(s,y). 6M

    Lesson 52. General Comments. Selecting h. Reducing h. Summary and an Example.

    A. Comment on Errors. 690 B. Choosing the Sise of h. 691 C. Re-ducing and Increasing h. 69B D. Summary and an Illustrative Exam-ple. 694-

    Lesson 53. Numerical Methods Applied to a System of Two

    690

    First Order Equations. 702

    Lesson 54. Numerical Solution of a Second Order Differential Equation. 707

    Lesson 55. Perturbation Method. First Order Equation. 713

    Lesson 56. Perturbation Method. Second Order Equation. 715

    ll. EXISTENCE AND UNIQUENESS THEOREM FOR THE FIRST ORDER DIFFERENTIAL EQUATION :r' = f(s,y). PICARD'S METHOD. ENVELOPES. CLAIRAUT EQUATION. 719

    Lesson 57. Picard's Method of Successive Approximations. 720

    Lesson 58. An Existence and Uniqueness Theorem for the First Order Differential Equation :r' ... f(s,y) Satisfying y(so) = :ro. 728

    A. Convergence and Uniform Convergence of a Sequence of Functions. Definition of a Continuous Function. 7B8 B. Lipschits Condition. Theorems from Analysis. 731 C. Proof of the Existence and Unique-ness Theorem for the First Order Differential Equation y' - /(z,y). 733

    Lesson 59. The Ordinary and Singular Points of a First Order Differential Equation y' - f(s,y). 744

  • CONTENTS xiii

    Lesson 60, Envelopes. 747 A. Envelopes of a Family of Curves. 7 48 B. Envelopes of a !-Param-eter Family of Solutions. 751,.

    Lesson 61. The Clairaut Equation. 757

    12. EXISTENCE AND UNIQUENESS mEOREMS FOR A SYSTEM OF FIRST ORDER DIFFERENTIAL EQUATIONS AND FOR LINEAR AND NONUNEAR DIFFERENTIAL EQUATIONS OF ORDER GREATER mAN ONE. WRONSKIANS. 763

    Lesson 62. An Existence and Uniqueness Theorem for a System of n First Order Differential Equations and for a Nonlinear Differential Equation of Order Greater Than One. 763

    A. The Existence and Uniqueness Theorem for a System of n First Order Differential Equations. 763 B. Existence and Uniqueness The-orem for a Nonlinear Differential Equation of Order n. 765 C. Exist-ence and Uniqueness Theorem for a System of n Linear First Order Equations. 768

    Lesson 63. Determinants. Wronskians. A. A Brief Introduction to the Theory of Determinants. 770 B. Wronskians. 771,.

    Lesson 64. Theorems About Wronskians and the Linear Independence of a Set of Solutions of a

    770

    Homogeneous Linear Differential Equation. 778

    Lesson 65. Existence and Uniqueness Theorem for the Linear Differential Equation of Order n. 783

    Bibliography 791

    Index 793

  • Preface for the Teacher

    IN WRITING THIS BOOK, it has been our aim to make it readable for the student, to include topics of increasing importance (such as transforms, numerical analysis, the perturbation concept) and to avoid the errors traditionally transmitted in an elementary text. In this last connection, we have abandoned the use of the terminology "general solution" of a differential equation unless the solution is in fact general, i.e., unless the solution actually contains every solution of the differential equation. We have also avoided the term "singular solution." We have exercised great care in defining function, differentials and solutions; in particular we have tried to make it clear that functions have domains.

    On the other hand, this accuracy has been secondary to our main pur-pose: to teach the student how to use differential equations. We hope and believe that we have not overlooked any of the major applications which can be made comprehensible at this elementary level. You will find in this text an extensive list of worked examples and homework problems with answers.

    We acknowledge our indebtedness to the publishers for their coopera-tion and willingness to let us use new pedagogical devices and to Prof. C. A. Hutchinson for his thorough editing.

    Ithaca, New York West Lafayette, Indiana

    XV

    M.T. H.P.

  • Preface for the Student

    THIS BOOK HAS BEEN WRITTEN primarily for you, the student. We have tried to make it easy to read and easy to follow.

    We do not wish to imply, however, that you will be able to read this text as if it were a novel. If you wish to derive any benefit from it, you must study each page slowly and carefully. You must have pencil and plenty of paper beside you so that you yourself can reproduce each step and equation in an argument. When we say '"verify a statement," "make a substitution," "add two equations," "multiply two factors," etc., you yourself must actually perform these operations. If you carry out the explicit and detailed instructions we have given you, we can almost guarantee that you will, with relative ease, reach the conclusion.

    One final suggestion-as you come across formulas, record them and their equation numbers on a separate sheet of paper for easy reference. You may also find it advantageous to do the same for Definitions and Theorems.

    Ithaca, New York West Lafayette, Indiana

    xvii

    M.T. H.P.

  • Chapter l

    Basic Concepts

    LESSON 1. How Differential Equations Originate.

    We live in a world of interrelated changing entities. The position of the earth changes with time, the velocity of a falling body changes with distance, the bending of a beam changes with the weight of the load placed on it, the area of a circle changes with the size of the radius, the path of a projectile changes with the velocity and angle at which it is fired.

    In the language of mathematics, changing entities are called variables and the rate of change of one variable with respect to another a deriva-tive. Equations which express a relationship among these variables and their derivatives are called differential equations. In both the natural and social sciences many of the problems with which they are concerned give rise to such differential equations. But what we are interested in knowing is not how the variables and their derivatives are related but only how the variables themselves are related. For example, from certain facts about the variable position of a particle and its rate of change with respect to time, we wish to determine how the position of the particle is related to the time so that we can know where the particle was, is, or will be at any time t. Differential equations thus originate whenever a uni-versal law is expressed by means of variables and their derivatives. A course in differential equations is then concerned with the problem of determining a relationship among the variables from the information given to us about themselves and their derivatives.

    We shall use an actual historical event to illustrate how a differential equation arose, how a relationship was then established between the two variables involved, and finally hpw from the relationship, the answer to a very interesting problem was determined. In the year 1940, a group of boys was hiking in the vicinity of a town in France named Lascaux. They suddenly became aware that their dog had disappeared. In the ensuing search he was found in a deep hole from which he was unable to climb out. When one of the boys lowered himself into the hole to help extricate the dog, he made a startling discovery. The hole was once a

    I

  • 2 BAsiC CoNCEPTS Chapter I

    part of the roof of an ancient cave that had become covered with brush. On the walls of the cave there were marvellous paintings of stags, wild horses, cattle, and of a fierce-looking black beast which resembled our bull.* This accidental discovery, as you may guess, created a sensation. In addition to the wall paintings and other articles of archaeological interest, there were also found the charcoal remains of a fire. The problem we wish to solve is the following: determine from the charcoal remains how long ago the cave dwellers lived.

    It is well known that charcoal is burnt wood and that with time cer-tain changes take place in all dead organic matter. It is also known that all living organisms contain two isotopes of carbon, namely C 12 and C14 The first element is stable; the second is radioactive. Furthermore the ratio of the amounts of each present in any macroscopic piece of living organism remains constant. However from the moment the organism dies, the C14 th!l-t is lost because of radiation, is no longer replaced. Hence the amount of the unstable C14 present in a dead organism, as well as its ratio to the stable C12, changes with time. The changing entities in this problem are therefore the element C14 and time. If the law which tells us how one of these changing entities is related to the other cannot be expressed without involving their derivative, then a differential equation will result.

    Let t represent the elapsed time since the tree from which the charcoal came, died, and let x represent the amount of C 14 present in the dead tree at any timet. Then the instantaneous rate at which the element C14 decomposes is expressed in mathematical symbols as

    (1.1)

    We now make the assumption that this rate of decomposition of C14 varies as the first power of X (remember X is the amount of 0 14 present at any timet). Then the equation which expresses this assumption is

    (1.11) dx -= -kx dt '

    where k > 0 is a proportionality constant, and the negative sign is used to indicate that x, the quantity of C14 present, is decreasing. Equation (1.11) is a differential equation. It states that the instantaneous rate of decomposition of C14 is k times the amount of C14 present at a moment of time. For example, if k = 0.01 and t is measured in years, then when x = 200 units at a moment in time, (1.11) tells us that the rate of decom-position of C14 at that moment is 1/100 of 200 or at the rate of 2 units per

    *You can see some of these pictures in PrimitiH Art by Erwin 0. Christensen, Viking Press, 1955, and in The Picture History of Painting by H. W. and D. J. Jansen, Harry N. Abrams, 1957.

  • Lesson I How DIFFERENTIAL EQUATIONS ORIGINATE 3

    year. If, at another moment of time, x = 50 units, then (1.11) tells us that the rate of decomposition of C14 at that moment is 1/100 of 50 or at the rate of ! unit per year.

    Our next task is to try to determine from (1.11) a law that will express the relationship between the variable x (which, remember, is the amount of C14 present at any time t) and the time t. To do this, we multiply (1.11) by dt/x and obtain (1.12)

    Integration of (1.12) gives (1.13)

    dx - = -kdt. X

    logx = -kt + c, where c is an arbitrary constant. By the definition of the logarithm, we can write (1.13) as (1.14) where we have replaced the constant e by a new constant A.

    Although (1.14) is an equation which expresses the relationship between the variable x and the variable t, it will not. give us the answer we seek until we know the values of A and k. For this purpose, we fall back on other available information which as yet we have not used. Since time is being measured from the moment the tree died, i.e., t = 0 at death, we learn from (1.14) by substituting t = 0 in it, that x = A. Hence we now know, since xis the amount of C14 present at any timet, that A units of C 14 were present when the tree, from which the charcoal came, died. From the chemist we learn that approximately 99.876 percent* of C 14 present at death will remain in dead wood after 10 years and that the assumption made after (1.1) is correct. Mathematically this means that when t = 10, x = 0.99876A. Substituting these values of x and t in (1.14), we obtain (1.15) 0.99876A = Ae-10k, 0.99876 = e-tok. We can now find the value of k in either or' two ways. There are tables which tell us for what value of -10k, e-tok = 0.99876. Division of this value by -10 will then give us the value of k. Or if we take the natural logarithm of both sides of (1.15) there results (1.2) log 0.99876 = -10k.

    *There is some difference among chemists in regard to this figure. The one used above is based on a half-life of Cl4 of 5600 years, i.e., half of Cl4 present at death will decompose in 5600 years. It is an approximate average of 5100 years, the lowest half-life figure, and 6200 years, the largest half-life figure.

  • 4 BASic CoNCEPTS Chapter 1

    From a table of natural logarithms, we find

    (1.21) -0.00124 = -10k, k = 0.000124 approximately. Equation (1.14) now becomes (1.22) where A is the amount of C14 present at the moment the tree died.

    Equation (1.22) expresses the relationship between the variable quan-tity ~ and the variable time t. We are therefore at last in a position to answer the original question,: How long ago did the cave dwellers live? By a chemical analysis of the charcoal, the chemist was able to determine the ratio of the amounts of C14 to C 12 present at the time of the dis-covery of the cave. A comparison of this ratio with the fixed ratio of these two carbons in living trees disclosed that 85.5 percent of the amount of C14 present at death had decomposed. Hence 0.145A units of C14 remained. Substit~ting this value for~ in (1.22}, we obtain (1.23) 0.145A = Ae-o.ooou4e

    0_145 = e-o.oool241

    log 0.145 = -0.000124t -1.9310 = -0.000124t

    t = 15573.

    Hence the cave dwellers lived approximately 15,500 years ago.

    Comment 1.3. Differential equation (1.11) originated from the as-sumption that the rate of decomposition of C 14 varied as the first power of the amount of C 14 present at any time t. The resulting relationship between the variables was then verified by independent experiment. Assumptions of this kind are continually being made by scientists. From the assumption a differential equation originates. From the differential equation a relationship between variables is determined, usually in the form of an equation. From the equation certain predictions can be made. Experiments must then be devised to test these predictions. If the pre-dictions are validated, we accept the equation as expressing a true law. It has happened in the history of science, because experiments performed were not sensitive enough, that laws which were considered as valid for many years were found to be invalid when new and more refined experi-ments were devised. A classical example is the laws of Newton. These were accepted as valid for a few hundred years. As long as the experi-ments concerned bodies which were macroscopic and speeds which were reasonable, the laws were valid. If the bodies were of the size of atoms or the speeds near that of light, then new assumptions had to be made, new

  • Lesson2A THE MEANING oF THE TERM Set 5

    equations born, new predictions foretold, and new experiments devised to test the validity of these predictions.

    Comment 1.4. The method we have described for determining the age of an organic archaeological remain is known as the carbon-14 test.*

    EXERCISE I 1. The radium in a piece of lead decomposes at a rate which is proporiional to

    the amount present. If 10 percent of the radium decomposes in 200 years, what percent of the original amount of radium will be present in a piece of lead after 1000 years?

    2. Assume that the half life of the radium in a piece of lead is 1600 years. How much radium will be lost in 100 years?

    3. The following item appeared in a newspaper. "The expedition used the carbon-14 test to measure the amount of radioactivity still present in the organic material found in the ruins, thereby determining that a town existed there as long ago as 7000 B.c." Using the half-life figure of C14 as given in the text, determine the approximate percentage of C14 still present in the organic material at the time of the discovery.

    ANSWERS I I. 59.05 percent. 2. 4.2 percent. 3. Between 32 percent and 33 percent.

    LESSON 2. The Meaning of the Terms Set and Function. Implicit Functions. Elementary Functions.

    Before we can hope to solve problems in differential equations, we must first learn certain rules, methods and laws which must be observed. In the lessons that follow, we shall therefore concentrate on explaining the meaning of certain terms which we shall use and on devising methods by which certain types of differential equations can be solved. We shall then apply these methods to solving a wide variety of problems of which the one in Lesson 1 was an example.

    We begin our study of differential equations by clarifying for you two of the basic notions underlying the calculus and ones which we shall use repeatedly. These are the notions of set and function.

    LESSON 2A. The Meaning of the Term Set. Each of you is familiar with the word collection. Some of you in fact may have or may have had collections-such as collections of stamps, of sea shells, of coins, of butter-flies. In mathematics we call a collection of objects a set, and the indi-

    *Dr. Willard F. Libby was awarded the 1960 Nobel Physics Prize for developing this method of ascertaining the age of ancient objects. His Cl4 half-life figure is 5600 years, the same as the one we used. According to Dr. Libby, the measurable age span by this test is from 1000 to 30,000 years.

  • 6 BAsic CoNCEPTs Chapter I

    vidual members of the set elements. A set therefore may be described by specifying what property an object must have in order to belong to it or by giving a list of the elements of the set.

    Examples of Sets. 1. The collection of positive integers less than 10 is a set. Its elements are 1, 2, 3, 4, 5, 6, 7, 8, 9.

    2. The collection of individuals whose surnames are Smith is a set. 3. The collection of all negative integers is a set. Its elements are

    ... '-4, -3, -2, -1. Since to each point on a line, there corresponds one and only one real

    number, called the coordinate of the point, and to each real number there corresponds one and only one point on the line, we frequently refer to a point on a line by its corresponding number and vice versa.

    Definition 2.1. The set of all numbers between any two points on a line is called an interval and is usually denoted by the letter I.

    If the two points on a line are designated by a and b, then the notation

    (2.11) I: a < x < b will mean the set of all real numbers x (or of all real values of x) which lie between the points a and b, but not including a and b. For convenience, we shall frequently omit the I and write only

    (2.111) a< x < b to represent this set of numbers. Similarly, (2.12) I: - oo < x < oo will mean the set ofall real values of x.

    I: a ~ x ~ b will mean the set of all real values of x between a and b, including the two end points.

    I: a ~ x < b will mean the set of all real values of x between a and b, including a but not b.

    I: -1 < x < 3, x = 10, will mean the set of all num-bers between -1 and 3 plus the number 10.

    I: x ~ 0 will mean the set of all positive real values of x plus zero.

    I: x = a will mean the set consisting of the single num-ber a.

    LESSON 2B. The Meaning of the Term Function of One Inde-pendent Variable. If two variables are connected in some way so that the value of one is uniquely determined when a value is given to the other, we say that one is a function of the other. (This concept will be given a more precise meaning in Definitions 2.3 and 2.31 below.)

  • Lesson2B THE MEANING OF THE TERM Function 7

    We shall show by examples below that the manner in which the rela-tionship between the variables is expressed is unimportant. It may be by an equation, of the kind with which you are familiar, or by other means. It is only important for the definition of a function that there be this unambiguous relationship between the variables so that, when a value is given to one, a corresponding value to the other is thereby uniquely determined.

    Example 2.2. Let l be the length of the side of a square and A its area. It is then customary to say that the area A depends on the length l, so that l is given an independent status and A a dependent one. How-ever, there is no valid reason why l could not be considered as being dependent on A. The decision as to which variable in a problem is to be considered as dependent and which independent lies entirely within the discretion of the individual. The choice will usually be determined by convenience. It is customary to write, whenever it is possible to do so, first the dependent variable, then an equals sign, then the independent variable in a manner which expresses mathematically the relationship be-tween the two variables.

    If in this example, therefore, we express the relationship between our two variables A and l by writing

    (a) we thereby give to A a dependent status and to l an independent one. Equation (a) now defines A as a function of l since for each l, it determines A uniquely. The relationship between the two variables, expressed mathe-matically by equation (a), is, however, not rigidly correct. It says that for each value of the length l, A, the area, is the square of l. But what if we let l = -3? The square of -3 is 9; yet no area exists if the side of a square has length less than zero. Hence we must place a restriction on l and say that (a) defines the area A as a function of the length l only for a set of positive values of land for l = 0. We must therefore write

    (b) l !1:; 0. Example 2.21. The relationship between two variables x and y is the

    following. If x is between 0 and 1, y is to equal 2. If x is between 2 and 3, y is to equal v'x. The equations which express the relationship between the two variables are, with the end points of the interval included,

    (a) y = 2, 0 ~X~ 1, Y = Vx, 2 ~ X ;:i! 3.

    These two equations now define y as a function of x. For each value of x in the specified intervals, a value of y is determined uniquely. The graph

  • 8 BASIC CoNCEPTs Chapter 1

    of this function is shown in Fig. 2.211. Note that these equations do not define y as a function of ~ for values of ~ outside the two stated intervals.

    y

    21---~

    (0,0) 2 3 X

    Figure 2.2ll

    The reason is obvious. We have not been told what this relationship is. For values of ~ therefore, equal to say i or -1 or 4, etc., we say that y is undefined or that the function is undefined.

    E:xample 2.22. In Fig. 2.221, we have shown the temperature T of a body, recorded by an automatic device, in a period of 24 consecutive hours. The horizontal axis represents the time in hours; the vertical axis the temperature T at any time t ~ 0. Even though we cannot express the relationship between th-e variables t and T by an equation, there can be no doubt that a precise, unique relationship between the two variables

    10 Th r J

    I I I~ (0,0) 3 6 9 412 ~1S-'2i'V24

    Figure 2.221

    exists. For each value of the time t, the graph will give a unique value of the temperature T. Hence the graph in this case defines T as a function oft.

    The main feature we wished to emphasize in the above differing exam-ples was that for the definition of a function it was not essential to be able to set up the relationship between the two variables by a single equation. (Most of the functions that you have encountered thus far were of this type.) As we mentioned at the outset, what is essential for the definition

  • Lesson 2B THE MEANING OF THE TERM Functwn 9

    of a function is that the relationship between the two variables be specific and unambiguous so that for each value taken on by an independent variable on a specified set, there should correspond one and only one value of a dependent variable. As you can verify, all the examples we gave above had this one common important property. We incorporate all the essential features of a function in the following definition.

    Definition 2.3. If to each value of an independent variable x on a set E (the set must be specified) there corresponds one and only one real value of the dependent variable y, we say that the dependent variable y is a function* of the independent variable x on the set E.

    It is customary to call the specified set E of values of the independent variable, the domain of definition of the independent variable and to call the set of resulting values of the dependent variable, the range of the dependent variable or the range of the function. Using this terminology, we may define a function alternately, as follows.

    Definition 2.31. A function is a correspondence between a domain set D and a set R that assigns to each element of D a unique element of R.

    Comment 2.32. A function is thus equivalent to a rule which tells us how to determine the unique element y of the range which is to be assigned to an element x of the domain. When we say therefore that the formula

    (a) y = v'X"=l, x 6; 1, defines y as a function of x, we mean that the formula has given us a rule by which, for each value of the independent variable x in its domain D: x 6; 1, we can determine the unique assigned value of yin the range R. (Here R 6; 0.) The rule is as follows: for each x in D, subtract one and take the positive square root of the result. For example, to the element x = 5 of D, the function or rule defined by (a) assigns the unique value 2 of R. Conversely, we can first give the rule, and then express the rule, if possible, by a formula.

    Comment 2.33. We may at times for convenience refer to an equa-tion or formula as if it were a function. For example, we may frequently refer to the equation

    (b) as a function. What we actually mean is that the equation y = x 2 defines a function; that is y = x2 gives us a rule by which to each x in D we can

    *In advanced mathematics, the function is called a real function. Since we shall for the most part be considering only real functions, we shall omit the word real whenever a real function is meant.

  • 10 BAsic CoNCEPTs Chapter I

    assign a yin R. Here the rule is: for each x in D, the number yin R assigned to it is obtained by squaring x.

    In view of Definition 2.3, such frequently encountered fonnulas as

    y = x2,

    y = v'1- x2, x2 + 5x

    y = X- 3

    are meaningless because they do not specify the domain, i.e., the set of values of x, for which the fonnulas apply. In practice, however, we inter-pret such formulas to define junctions for all values of x for which they make sense. The first equation, therefore, defines a function for all values of x, the second for values of x in the interval -1 ~ x ~ 1, and the last for all values of x except x = 3.

    Because of the above comment, the function defined by

    y = v'1- x2 is not the same as the function defined by

    y = v'1 - x2, 0 ~ x ~ 1. (The first is defined for all x in -1 ~ x ~ 1; the second only for x in 0 ~ X ~ 1.)

    It is customary and convenient to record the fact that the dependent variable y is a function of the independent variable x (these are the letters most commonly used for the dependent and independent variable respec-tively), by means of the symbolic expression (2.34) y = f(x). It is read as "y equals f of x" or "y is a function of x."

    By Definition 2.3 and (2.34), we could therefore write for the tempera-ture Example 2.22,

    (2.35) T = j(t). We then say T is a function of t and refer to the graph itself as the defini-tion or rule which tells us which T to assign to each value of t.

    Comment 2.36. We shall at times write an expression in x, say

    (a} f(x) = x2 + es, - oo < x < oo and refer to f(x) as a function. What we mean is that f(x), here (x2 + es), gives a rule by which for each x, we can assign a unique value to f(x).

  • Lesson2C FuNCTION OF Two INDEPENDENT VARIABLES 11

    Definition 2.4. If f(x) is a function of x defined on a set E, then the symbolf(a), for any a in E, means the unique value assigned to f(x) ob-tained by substituting a for x.

    E:~tample 2.5. If (a) f(x) = x2 + 2x + 1, 0 ;:i x ;:i 1, find f(O), j(1), j(i), !(2), f( -1).

    Solution. By Definition 2.4 we have

    (b) j(O) = 02 + 2 0 + 1 = 1, !(1) = 12 + 2 . 1 + 1 = 4, j(i) = (!) 2 + 2 . i + 1 = 2:1, f(2) is undefined since 2 is not in our set E: 0 ;:i x ;:i 1, f( -1) is undefined since -1 is not in our set E.

    If several functions appear in a single context so that the use of the same letter for each would be confusing, it is permissible to replace f by other letters. Those most frequently used are g, h, G, H, F, etc. Similarly, we may use other letters in place of x andy. Those usually used are the ones at the end of the alphabet, namely u, v, w, z, s, t.

    LESSON 2C. Function of Two Independent Variables. In Lesson 2B, we defined a function of one independent variable. In an analogous manner, we define a function of two independent variables x and y as follows.

    Definition 2.6. If to each element (x,y) of a set E in the plane (the set must be specified) there corresponds one and only one real value of z, then z is said to be a function of x and y for the set E. In this event, x,y are called independent variables and z a dependent variable.

    As in the case of one independent variable, the set E is sometimes called the domain of definition of the function and the set of resulting values of z, the range of the function. '

    In view of Definition 2.6, a formula such as

    is meaningless since it does not specify the domain of definition. Here again as in the case of one variable, we interpret such formulas to define functions for all values of x and y for which they make sense. In this example therefore the elements of the domain D are the points (x,y) in the plane where -1 ;:i x ;:i 1, - oo < y < oo. The domain, therefore, for which the formula defines z as a function of x and y consists of all points

  • 12 BAsic CoNCEPTs Chapter 1

    in the plane between and including the lines x = 1 and x = -1. It is the shaded area in Fig. 2.61.

    z

    y

    Figure 2.61

    E%ample 2.62. Determine the domain D for which each of the fol-Jowing formulas define z as a function of x and y.

    v'l=Y2 1. z = . v'f"=X2

    2. Z =X+ y. 3. z = vx2 + y2 - 25. 4. z = v-(x2 + y2). 5. z.= v-(x2 + y2 + 1).

    Solutions. 1. The elements of the domain D for which the formula defines z as a function of x and y, consists of those points (x,y) where -1 < x < 1, -1 ;:i! y ;:i! 1. The domain D is the shaded square shown

    z z

    y y

    (a) (b) Figure2.63

    in Fig. 2.63{a). It is bounded by the lines x = 1 and y = 1. It includes the lines y = 1 but not the lines x = 1.

    2. The domain is the entire plane.

  • Lesson 2C FuNcriON OF Two INDEPENDENT VARIABLES 13

    3. The domain consists of those points (x,y) for which x2 + y 2 ~ 25. It is the shaded area in Fig. 2.63(b), i.e., it is the area outside the circle x2 + y2 = 25 plus the points on its circumference.

    4. The domain consists of the single point (0,0). 5. This formula does not define z as a function of x andy. There does

    not exist a domain which will determine a value of z. It is evident from the above examples, that a two-dimensional domain

    may cover the whole plane or part of the plane; it may cover the whole plane with the exception of a hole in its interior; its boundaries may be circular or straight lines, or it may consist of only a finite number of points. In short, in contrast to a one-dimensional domain, a two-dimensional domain may assume a great variety of shapes and figures.

    It is customary and convenient to record the fact that z is a function of x and y by means of the symbolic expression

    (2.64) z = f(x,y). It is read as "z equals f of x,y" or "z is a function of x,y."

    Definition 2.65. If f(x,y) is a function of two independent variables x,y, defined over a domain D, then the symbol f(x,a) for any element (x,a) in D, means the function of x obtained by replacing y by a.

    Example 2.66. If

    (a) f(x,y) = x 2 +xy2 +5y+3, -oo < x < oo, -oo < y < oo, find f(x,2), f(x,a), f[x,g(x)].

    Solution. By Definition 2.65

    (b) f(x,2) = x2 + 4x + 10 + 3 = x2 + 4x + 13, -oo < x < oo, f(x,a) = x2 + a2x + 5a + 3, - oo < x < oo.

    And, if g(x) is defined for all x,

    f[x,g(x)] = x2 + x[g(x)]2 + 5g(x) + 3, -oo < x < oo. Example 2.67. If

    (a) f(x,y) = x + y, -1 ~ x ~ 1, 0 ~ y ~ 2, find f(x,!), f(x,3).

    Solution. By Definition 2.65

    (b) f(x,!) = X+!, -1 ;:i! X ~ 1. But f(x,3) is undefined since the domain of y is the interval 0 ;:i! y ;:i! 2.

    A special type of set, called a region, is defined as follows.

  • 14 BASIC CONCEPTS Chapter 1

    Definition 2.68. A set in the plane is called a region if it satisfies the following two conditions: 1. Each point of the set is the center of a circle whose entire interior con-

    sists of points of the set. 2. Every two points of the set can be joined by a curve which consists

    entirely of points of the set.

    In Example 2.62, the domain defined in 2 is a region. If the boundary points are excluded from each set defined in 1 and 3, then each resulting domain is also a region. Each point of each set satisfies requirement 1, and every two points of eacp set satisfies requirement 2. On the other hand, the set consisting of the points on a line is not a region. The set satisfies requirement 2 but not 1. Also the set consisting of isolated points is not a region-the points in the set do not satisfy either of the two re-quirements.

    Definition 2.69. A region is said to be bounded if there is a circle which will enclose it.

    Comment 2.691. In a manner analogous to Definition 2.6, we can define a function of three or more independent variables.

    LESSON 2D. Implicit Function. Consider a relationship between two variables x,y given by the formula

    (2.7) Does it define a function? If x > 5 or x < -5, then the formula will not determine a value of y. For example, if x = 7, there is no value of y which will make the left side of (2.7) equal to zero. (Why?) However, if x lies between -5 and 5 inclusive, then there is a value of y which will make the left side of (2.7) equal to zero. To find it, we solve (2.7) for y and obtain (2.71) y = v'25 - x2, -5 ~ x ~ 5. When the relation between x and y is written in this form, however, we see that the formula does not define y uniquely for a value of x. Hence, by our Definition 2.3, it does not define a function. We can correct this defect by specifying which value of y is to be chosen. For example, we can choose any one of the following three formulas to determine y.

    {2.72) y = v'25- x2, -5 ~X~ 5.

    {2.73) y = -v'25 - x2, -5 ~X~ 5. {2.74) y = v'25- x2,

    -5 ~X~ 0; = -v'25- x2, 0

  • Le810n2D IMPLICIT FuNCTION IS

    By Definition 2.3, each of these formulas now defines a function. It gives a rule which assigns a unique y to each x on the specified interval.

    Now consider the formula

    (2.75) which also connects two variables x andy, and ask of it the same ques-tion. Does it define a function? If it does, then there must be values of x for which it will determine uniquely values of y. It should be evident to you that there are no values of y for any x. (Write the equation as x2 + y2 = -1.) Hence this equation does not define a function by our Definition 2.3.

    As a final example, we consider the formula

    (2.76) and again ask the question. Does it define a function? And if it does, for what values of x will it determine uniquely a value of y? The answer to both questions, unlike the answer to the previous formula (2.7}, is not easy to give. For unlike it, (2.76) cannot be solved easily for yin terms of x. Hence we must resort to other means. The graph of equation (2.76} is shown in Fig. 2.77.

    Figure 2.77

    From the graph, we see that for x ;::i! 0 and x > 22' 3, y is uniquely determined. Hence formula (2.76) does define y as a function of x in these two intervals, but not in the interval 0 < x ;::i! 22' 3 It is possible, however, to make formula (2.76) define y as a function of x for all x if we

  • 16 BASIC CONCEPTS Chapter I

    choose one of the three possible values of y for each x in the interval 0 < x < 22' 3 and choose one of the two possible values of y for x = 22' 3 With these restrictions, we then would be able to assert that (2.76) defines y as a function of x for all x.

    Whenever a relationship which exists between two variables x and y is expressed in the form (2.7) or (2.76), we write it symbolically as (2.8) f(x,y) = 0. It is read as ''! of x,y equals zero," or as "a function of x,y equals zero."

    If the relation which defines y as a function of x is expressed in the form f(x,y) = 0, it is customary to call y an implicit function of x. When we say therefore that y is an implicit function of x, we mean, as the name suggests, that the functional relationship between the two variables is not explicitly visible as when we write y = f(x), but that it nevertheless im-plicitly exists. That is, there is a function, let us call it g(x), which is implicitly defined by the relationf(x,y) = 0 and which determines uniquely a value of y for each x on a set E. Hence:

    Definition 2.81. The relation (2.82) f(x,y) = 0

    defines y as an implicit function of x on an interval I: a < x < b, if there exists a function g(x) defined on I such that

    (2.83) j[x,g(x)] = 0 for every x in I.

    E:tample 2.84. Show that

    (a) f(x,y) = x 2 + y 2 - 25 = 0 defines y as an implicit function of x on the interval I: -5 ;:i! x ;:i! 5.

    Solution. Choose for g(x) any one of the functions (2.72), (2.73), or (2.74). If, for example, we choose (2.72), then g(x) = y25 - x2. It is defined on I, and by (a) and Definition 2.65, (b) f[x,g(x)] = x2 + [v25 - x2]2 - 25 = o. Hence Definition 2.81 is satisfied.

    E:tample 2.85. Show that

    (a) f(x,y) = x 3 + y 3 - 3xy = 0 defines y as an implicit function of x for all x.

  • Leaeon2E THE ELEMENTARY FuNCTIONS 17

    Solution.. Here, as pointed out earlier, it is not easy to solve for y in terms of :e, so that it is not easy to find the required function g(:e). How-ever, if we select from Fig. 2.77 any one of the graphs shown in Fig. 2.86

    y

    Figure 2.86

    to represent the function g(:e), then j[:e,g(:e)] = 0 for every :e. Hence Definition 2.81 is satisfied. (Note that g(z).may be selected in infinitely many more ways.) LESSON 2E. The Elementary Functions. In addition to the tenns function and implicit function, we shall refer at times to a special class of functions called the elementary functions. These are the constants and the following functions of a variable :e: 1. Powers of x: x, x2, x3, etc. 2. Roots of x: vx, -X, etc. 3. Exponentials: e"'. 4. Logarithms: log x. 5. Trigonometric functions: sin x, cos x, tan x, etc. 6. Inverse trigonometric functions: Arc cos x, etc. 7. All functions obtained by replacing x any number of times by any of

    the other functions 1 to 6. Examples are: log sin x, sin (sin x), eoins, easl, etc.

    8. All functions obtained by adding, subtracting, multiplying, and divid-ing any of the above seven types a finite number of times. Examples

    ein., es2+2s+l are: 2x - log x + -x-, (Arc cos x) 2 + V3x' - log (log 4x). In the calculus course, you learned how to differentiate elementary

    functions and how to integrate the resulting derivatives. If you have for-gotten how, it would be an excellent idea at this point to open your cal-culus book and review this material.

  • 18 BAsic CoNCEPTS Chapter!

    EXERCISE 2 1. Describe, in words, each of the following sets:

    (a) X < 0. (b) X ;:i! 0. (c) a < X ;:i! b. (d) -oo < X < 5, X - 7, (e) -3 < X < -2, X > 0. (f) V2 < X < 11'. (g) 211' ;:i! X < 311',

    2. Define the area A of a circle as a function of its radius r. Which is the de-pendent variable and which is the independent variable? Draw a rough graph which will show how A depends on r, when r is given values between 0 and 5.

    3. Under certain circumstances the pressure p of a gas and its volume V are related by the formula p V312 = 1. Express each variable as a function of the other. '

    4. Explain the difference between the function

    y=vx. 2;:iix;:ii3, and the function

    y = Vz1 X> 0. 5. Let

    F(x) = 2 ifx < 0, = 7 if 0 ~ X ;:i! 1, =x3 -2x ifx>4.

    Find (a) F(-1), (b) F(O), (c) F(0.7), (d) F(4), (e) F(S), (0 F(2). 6. If

    x2 - 2x + 1 g(x) = 1 1 X ~ 1, x-find (a) g(2), (b) g(-5), (c) g(1), (d) g(u), (e) g(x2), (0 g(x- 1).

    7. Why is the function defined in 6 not the same as the function defined by g(x) = x- 1?

    8. Determine the domain D for which each of the following formulas defines z as a function of x and y.

    v'f=X2 (a) z = v'l=7i2

    (c) z = Vx2 + (y- 1)2. (e) z = ...j -(x + y)2. (f) z = V-(x2 + y2 + 3).

    (b) Z = X - y + 2.

    (d) z = Vx2+ y2- 9.

    1-x (g) z = -- 1-y 9. Which of the domains in problem 8 are regions?

    10. If f(x,y) is a function of two independent variables x,y defined over a domain D, then the symbol f(b,y) for any element (b,y) in D, means the function of y obtained by replacing x by b; see Definition 2.65. Let

    f(x,y) = x2 + 2xy + log (xy), xy > 0. Find: (a) f(x,1), f(x,b), f(x,O), f(x,x2). (b) f(1,y), f(a,y), J(O,y), f(x2,y). (c) /(1,1), /(2,-1), J(a,b), J(u,v).

  • Lesson 2-Exercise

    11. Draw the graphs of three different functions defined by x3 + y3 - 3xy = 0.

    See Fig. 2.77. Is there one which is continuous for all x? 12. The following is a standard type of exercise in the calculus.

    19

    dy dy If x3 + y3 - 3xy = 0, then 33:2 + 3y2 dx - 3x dx - 3y = 0. Therefore

    dy y- x2 2 -=---, y r'x d:z: y2- X

    Explain by the use of Fig. 2.77 what this means geometrically. 13. Explain why the procedure followed in problem 12, applied to the relation

    x2 + y2 + 1 = 0 and yielding the result

    is meaningless.

    dy X dx=-y

    14. Find the function g(x) that is implicitly defined by the relation

    Vx2 - y2 +Arc cos! = 0 y r' 0 y ' '

    15. Explain why Vx2 -y2+Arcsin! = 0 y

    does not define y as an implicit function of x: 16. Can you apply the method of implicit differentiation as taught in the calculus

    to the function of problem 14, of pr.oblem 15? 17. Define a function of three independent variables x1, x2, x3; of n independent

    variables x1 , x,.. Hint. See Definition 2.6.

    ANSWERS 2 1. (a) The set of all negative values of x. tb) The set (a) plus zero.

    (c) The set of values of x between a and b, including b but not a. (d) The set of all values of x less than five, plus the number 7. (e) The set of all values of x between -3 and -2, plus all positive values of x.

    2. A = rr2, r ~ 0. 3. P = v-312, v > o; v = p-213, P > o. 4. The first function is defined only for all x between and including two and

    three; the second function is defined for all x greater than zero. 5. (a) 2. (b) 7. (c) 7. (d) Undefined. (e) 496. (f) Undefined. 6. (a) 1. (b) -6. (c) Meaningless. (d) (u2 - 2u + 1)/(u- 1).

    u r' 1. (e) (x4 - 2x2 + 1)/(x2 - 1), x2 ;o' 1. (f) (x2 - 4x + 4)/ (x - 2), x r' 2.

    7. Function defined in 6 if! meaningless when x = 1. 8. (a) -1 ~ x ;:;i; 1, -1 < y < 1. (b) Entire plane. (c) Entire plane.

    (d) Area outside circle x2 + y2 = 9 plus points on the circumference of the circle. (e) Line x + y = 0. (f) Nonexistent. (g) y r' 1.

    9. (b), (c), (g). Also (a) and (d) if their boundary points are excluded. 10. (a) x2 + 2x + log x; x2 + 2bx + log (bx) ; undefined; x2 + 2x3 + log x3.

    (b) 1 + 2y +logy; a2 + 2ay +log ay; undefined; x4 + 2x2y +log (x2y). (c) 3; undefined; a2 + 2ab +log (ab); u2 + 2uv +log (uv).

  • 20 BAsic CoNCEPTS Chapter 1

    13. x2 + y2 + 1 = 0 does not define a function. 14. y = g(x) = x. 15. The first term requires that I x I e; I y I ; the second term that I x I ~ I y I . 16. No, both examples.

    LESSON 3. The Differential Equation.

    LESSON 3A. Definition of an Ordinary Differential Equation. Order of a Differential Equation. In the calculus, you studied various methods by which you could differentiate the elementary functions. For example, the successive derivatives of y = log x are

    (a) 1 y' =-I X

    -1 ylt = -2 I

    X 2

    y'" = z3 1 etc.

    And if z = x3 - 3xy + 2y21 its partial derivatives with respect to x and with respect to y are respectively

    oz 2 (b) ox = 3x - 3y1 az -= -3x+4y ay I Equations such as (a) and (b) which involve variables and their deriva-

    tives are called differential equations. The first involves only one inde-pendent variable x; the second two independent variables x andy. Equa-tions of the type (a) are called ordinary differential equations; of the type (b) partial differential equations. Hence,

    Definition 3.1. Let f(x) define a function of x on an interval 1: a < x < b. By an ordinary differential equation we mean an equa-tion involving x, the function f(x) and one or more of its derivatives.

    Note. It is the usual custom in writing differential equations to replace

    f(x) by y. Hence the differential equation d';) + x[f(x)] 2 = o is usually

    written as:+ xy2 = 0; the differential equation D.,2[J(x)] + xDJ(x) = ez as Dz 2y + xDzY = ez or as y" + xy' = ez.

    Examples of ordinary differential equations are:

    (3.11) :+Y = 0. (3.12) y' = ez.

    (3.13) d2y 1 dx2 = 1- x2 (3.14) f'(x) = f"(x). (3.15) xy' = 2y.

  • Lesson 3B

    (3.16) (3.17) (3.18)

    ExPLICIT SoLUTION OF A DIFFERENTIAL EQuATION 21

    y'' + (3y') 3 + 2x = 7. (y"')2 + (y")4 + y' = X.

    xy + 2y" + (xy') 5 = x3 Note. Since only ordinary differential equations will be considered in

    this text, we shall hereafter omit the word ordinary.

    Definition 3.2. The order of a differential equation is the order of the highest derivative involved in the equation.

    For the differential equations listed above, verify that (3.11), (3.12), and (3.15) are of the first order; (3.13), (3.14), and (3.16) are of the second order; (3.17) is of the third order; (3.18) is of the fourth order.

    A WORD OF CAUTION. You might be tempted to assert, if you were not careful, that

    y"-y"+y'-y=O

    is a second order differential equation because of the presence of y". How-ever, y" is not really involved in the equation since it is removable. Hence the equation is of order 1.

    LESSON 3B. Solution of a Differential Equation. Explicit Solu-tion. Consider the algebraic equation

    (3.3) x2 - 2x- 3 = 0. When we say x = 3 is a solution of (3.3), we mean that x = 3 satisfies it, i.e., if xis replaced by 3 in (3.3), the equality will hold. Similarly, when we say the function f(x) defined by (3.31) y = f(x) = logx + x, x > 0, is a solution of

    (3.32) x 2y" + 2xy' + y = log x + 3x + 1, x > 0, we mean that (3.31) satisfies (3.32), i.e., if in (3.32) we substitute the function f(x) = log x + x for y, and the first and second derivatives of the function for y' and y", respectively, the equality will hold. [Be sure to verify the assertion that (3.31) does in fact satisfy (3.32).]

    We want you to note two things. First in accordance with Definition 2.3, we specified in (3.31) the values of x for which the function is defined. But even if we had not, the interval x > 0 would have been tacitly assumed sincfl log x is undefined for x ;:;;! 0. Second, we also specified in (3.32) the interval for which the differential equation makes sense. Since it too contains the term log x, it too is meaningless when x ;:;;! 0.

  • 22 BASic CoNCEPTs Chapter 1

    Definition 3.4. Let y = j(x) define y as a function of x on an interval I: a < x < b. We say that the function j(x) is an explicit solution or simply a solution of an ordinary differential equation involving x, j(x), and its derivatives, if it satisfies the equation for every x in I, i.e., if we replace y by j(x), y' by f'(x), y" by f"(x), , y by tn>(x}, the differen-tial equation reduces to an identity in x. In mathematical symbols the definition says: the function j(x) is a solution of the differential equation (3.41) if (3.42) for every x in I.

    F(x,y,y', , y) = 0,

    F[x,f(x},f'(x), ,j(x)] = 0

    Comment 3.43. We shall frequently use the expression, "solve a dif-ferential equation," or "find a solution of a differential equation." Both are to be interpreted to mean, find a junction which is a solution of the differential equation in accordance with Definition 3.4. Analogously when we refer to a certain equation as the solution of a differential equation, we mean that the junction defined by the equation is the solution. If the equa-tion does not define a function, then it is not a solution of any differential equation, even though by following a formal procedure, you can show that the equation satisfies the differential equation. For example, the equation y = v'-(1 + x2) does not define a function. To say, therefore, that it is a solution of the differential equation x + yy' = 0 is meaningless even though the formal substitution in it of y = v'- (1 + x2) and y' = -x/v'-(1 + x2) yields an identity. (Verify it.)

    Example 3.5. Verify that the function defined by (a) is a solution of the differential equation (b) (y") 3 + (y') 2 - y - 3x2 - 8 = 0.

    Solution. By (a), the function j(x) = x2 Therefore f'(x} = 2x, f"(x) = 2. Substituting these values in (b) for y, y', y", we obtain (c) 8 + 4x2 - x2 - 3x2 - 8 = 0. Since the left side of (c) is zero, (a), by Definition 3.4, is an explicit solu-tion or simply a solution of (]:>). Note that (b) is also defined for all x.

    Remark. It is the usual practice, when testing whether the function defined by the relation y = j(x) on an interval I is a solution of a given differential equation, to substitute in the given equation the values of y and its derivatives. In the previous Example 3.5, therefore, if we had followed this practice, we would have substituted in (b): y = x2, y' = 2x, y" = 2. If an identity resulted, we would then say that (a) is a solution

  • Lenon3B ExPLICIT SoLUTION oF A DIFFERENTIAL EQUATION 23

    of (b). We shall hereafter, for convenience also follow this practice, but you should always remember that it is the functionf(x) and its derivatives which must be substituted in the given differential equation for y and its corresponding derivatives. And if y = f(x) does not define a function then y or /(x) cannot be the solution of any differential equation.

    Example 3.51. Verify that the function defined by (a) is a solution of

    (b)

    y = log X + C1 X > 0

    1 y' = -

    X

    Solution. Note first that (b) is also defined for all x > 0. By (a), y' = 1/x. Substituting this value of y' in (b) gives an identity. Hence (a) is a solution of (b) for all x > 0.

    Example 3.52. Verify that the function defined by

    (a) y = tan x - x, 7r X F- (2n + 1) 2 I n = 01 11 21 ' ' I is a solution of

    (b) Solution. Here y = tan x - x, y' = sec2 x - 1 = tan2 x. Substi-

    tution of these values in (b) for y andy' gives the identity (c) tan2 x = (x + tan x - x) 2 = tan2 x. Hence (a) is a solution of (b) in each of the intervals specified in (a).

    Comment 3.521. Note by (b) that the differential equation is defined for all x. Its solution, however, as given in (a), is not defined for all x. Hence, the interval, for which the function defined in (a) may be a solu-tion of (b), is the smaller set of intervals given in (a).

    Comment 3.53. It is also pos-sible for a function to be defined over an interval and be the solution of a differential equation in only part of this interval. For example y = lxl is defined for all x. Its graph

    y

    is shown in Fig. 3.54. It has no de- x rivative when x = 0. It satisfies the differential equation y' = 1 in Figure 3.54 the interval x > 0, and the differ-ential equation y' = -1 in the interval x < 0. But it does not satisfy any differential equation in an interval which includes the point x = 0.

  • 24. BASIC CoNCEPTS Chapter 1

    LESSON 3C. Implicit Solution of a Dift'erential Equation. To test whether an implicit function defined by the relation f(x,y) = 0 is a solution of a given differential equation, involves a much more compli-cated procedure than the testing of one explicitly expressed by y = f(x). The trouble arises because it is usually not easy or possible to solve the equation f(x,y) = 0 for y in terms of x in order to obtain the needed function g(x) demanded by Definition 2.81. However, whenever it can be shown that an implicit function does satisfy a given differential equation on an interval I: a < x < b, then the relation f(x,y) = 0 is called (by an unfortunate usage)* an implicit solution of the differential equation.

    Definition 3.6. A relation f(x,y) = 0 will be called an implicit solu-tion of the differential equation

    (3.61) F(x,y,y', 1 y) = 0 on an interval I: a < x < b, if 1. it defines y as an implicit function of x on I, i.e., if there exists a func-

    tion g(x) defined on I such that f(x,g(x)] = 0 for every x in I, and if 2. g(x) satisfies (3.61), i.e., if (3.62) F[x,g(x),g'(x), , g(x)] = 0 for every x in I.

    Emmple 3.63. Test whether

    (a) f(x,y) ;,. x2 + y 2 - 25 = 0 is an implicit solution of the differential equation

    (b) F(x,y,y') = yy' + x = 0 on the interval I: -5 < x < 5.

    Solution. We have already shown that (a) defines y as an implicit function of x on I, if we choose for g(x) any one of the functions (2.72), (2.73), (2.74). If we choose (2.72), then (c) g(x) = v'25 - x2, g'(x) = - x ; -5 < x < 5.

    v'25- x 2

    Substituting in (b), g(x) for y, g'(x) for y', there results (d) F[x,g(x), g'(x)] = v'25 - x2 (- x ) + x = 0.

    v'25- x2 ------

    Actually /(z,y) = 0 is an equation and an equation is never a solution of a differential equation. Only a junction can be a solution. What we really mean when we say /(z,y) = 0 is a solution of a differential equation is that the junction g(z) defined by the relation /(z,y) = 0 is the solution. See Definition 3.6, also Comment 3.43.

  • Lesson 3C IMPLICIT SOLUTION OF A DIFFERENTIAL EQUATION 25

    Since the left side of (d) is zero, the equation is an identity in x. There-fore, both requirements of Definition 3.6 are satisfied, and (a) is therefore an implicit solution of (b) on I.

    Example 3.64. Test whether

    (a) f(x,y) = x3 + y3 - 3xy = 0, -ao < x < ao, is an implicit solution of

    (b) F(x,y,y') = (y2 - x)y' - y + x2 = 0, -ao < x < ao. Solution. Unlike the previous example, it is not easy to solve for y

    to find the required function g(x). However, we have already shown that (a) defines y as an implicit function of x, if we choose for g(x) any one of the curves in Fig. 2.86. (Be sure to refer to the graphs shown in this figure.) If we choose the first, then g'(x) does not exist when x = 22' 3 Hence, (a) cannot be a solution of (b) for all x, but we shall show that (a) is a solution of (b) in any interval which excludes this point x = 22' 3 Since we do not have an explicit expression for g(x), we cannot substitute in (b), g(x) for y, g'(x) for y' to determine whether F[x,g(x), g'(x)] = 0. What we do is to differentiate (a) implicitly t? obtain (c) 3x2 + 3y2y' - 3xy' - 3y = 0, (y2 - x)y' - y + x 2 = 0. Since (c) now agrees with (b), we know that the slope of the function g(x) implicitly defined by (a) and explicitly defined by the graph, satisfies (b) at every point x in any interval excluding x = 2213 Hence both re-quirements of Definition 3.6 are satisfied in any interval which does not include the point x = 22' 3 Therefore the function g(x) defined by the first graph in Fig. 2.86 is an implicit solution of (a) in any interval not containing the point x = 22' 3

    If we choose for g(x) the second curve in Fig. 2.86, then g'(x) does not exist when x = 0 and x = 22' 3 For this g(x), (a) will be a solution of (b) in any interval which excludes these two points.

    If we choose for g(x) the third curve in Fig. 2.86, then g'(x) does not exist when x = 0. For this g(x), (a) will be a solution of (b) in any inter-val which excludes this point.

    Comment 3.65. The example above demonstrates the possibility of an implicit function being defined over an interval and being the solution of a differential equation in only part of the interval.

    Comment 3.651. The standard procedure in calculus texts to prove that (a) is a solution of (b) is the following. Differentiate (a) implicitly. If it yields (b), then (a) is said to be an implicit solution of (b). If you operate blindly in this manner, then you are likely to assert that x2 +

  • 26 BASIC CONCEPTS Chapter 1

    y2 = 0 is an implicit solution of x + yy' = 0, since differentiation of the first gives the second. But x2 + y2 = 0 does not define y implicitly as a function of x on an interval. Only the point (0,0) satisfies this fonnula. To assert, therefore, that x2 + y 2 = 0 is an implicit solution of x + yy' = 0 because it satisfies the differential equation is meaningless.

    Example 3.66. Test whether

    (a) xy2 - e_, - 1 = 0 is an implicit solution of the differential equation

    (b) (xy2 + 2xy- 1)y' + y2 = 0. Solution. If we worked blindly and used the method of implicit dif-

    ferentiation as taught in the calculus, then from (a), we would obtain by differentiation

    (c) Although (c) is not identical with (b), it can be made so if we replace e-11 in the second equation of (c) by its value xy2 - 1 as determined from (a). Working blindly then, we would assert that (a) is an implicit solu-tion of {b). But is it? Well, let us see. By Definition 3.6, we must first show that (a) defines y as an implicit function of x on an interval. If we write (a) as (d) ~1 +e-ll y.= ,

    X

    y

    X

    Figure 3.67

    we see, since e-11 is always positive, that y is defined only for x > 0. Hence the interval for which (a) may be a solution of (b) must exclude

  • Lesson 3-Exercise 27

    values of x ~ 0. Here again as in the previous example, we cannot easily solve for y explicitly in terms of x, so that we must resort to a graph to determine g(x). It is given in Fig. 3.67. From the graph, we see that there are three choices of g(x). If g(x) is the upper branch, then (a) is an implicit solution of (b) for all x > 0. And if g(x) is either of the two lower branches, one above the line y = -2.22, the other below this line, then (a) is an implicit solution of (b) for I: x > 2.07 approximately.

    EXERCISE 3

    1. Determine the order of the following differential equations. (a) dy + (xy - cos x) dx = 0. (b) y" + xy" + 2y(y')3 + xy = 0. (c) (~~r - (y111)4 +X = 0. (d) et/" + xy11 + y = 0.

    2. Prove that the functions in the right-hand column below are solutions of the differential equations in the left-hand columns. (Be sure to state the common interval for which solution and differential equation make sense.) (a) y' + y = 0 y = e-"'. (b) y' = e"' y = e"'.

    d2y 1 (c) d:J:2 = v'f=X2 (d) J' (x) = !" (x) (e) xy' = 2y (f) (1 + x2)y' = xy (g) cos8 :; - 2rsin8 = 0 (h) y"- y = 0 (i) J' (x) = !J(x) (j) xy' + Y = Y2 (k) X+ yy' = 0

    y = x Arc sin x + v'f=X2. y = e"'+ 2. Y = x2. y = ~1 + x2.

    y = ae"' + be-"'. J(x) = 26"13

    2 y=-- x+2

    y = ~16- x2. 3. Show that the differential equation

    I :: I + IYI + 1 ..; 0 has no solutions.

    4. Determine whether. the equations on the right define implicit functions of x. For those which do, determine whether they are implicit solutions of the differential equations on the left. (a) y2 - 1 - (2y + xy)y' = 0 y2 - 1 = (x + 2) 2 (b) e"'-" + e"-"' dy = 0 i 11 + i"' = 1. dx (c) dy = - '!1. x2 + l + 1 = 0. dx X

  • 28 BABIC CoNCEPTS Chapter 1

    ANSWERS 3 1. (a) 1. (b) 2. (c) 3. (d) 3. 2. (a) -co < x < co,

    (d) -co < x < co. r 3r

    (g) 8 ;>(b) -co < x < co. (e) x 0.

    (c) -1 < x < 1. (f) -co < x < co.

    ...

    (i) -co < x < co. (j) X;>4. (a) Yes, if one makes y single valued. Implicit solution. (b) Yes. Implicit solution, x 0. (c) Function undefined.

    LESSON 4. The General Solution of a Differential Equation.

    LESSON 4A. Multiplicity of Solutions of a Differential Equation. We assume at the outset that you have understood clearly the material of the previous lesson so that when we say "solve a differential equation" or "find a solution of a differential equation," or "the solution of a differen-tial equation is," you will know what is meant (see Comment 3.43). Or if we omit intervals for which a function or a differential equation is defined, we expect that you will be able to fill in this omission yourself.

    When you studied the theory of integration in the calculus, you solved some simple differential equations of the form y' = f(x). For example, you learned that, if

    (4.1) y' = e"', then its solution, obtained by a simple integration, is

    (4.11) y=e"'+c, where c can take on any numerical value. And if

    (4.12) y" = e"', then its solution, obtained by integrating (4.12) twice, is (4.13) where now c1 and c2 can take on arbitrary values. Finally, if

    (4.14) y"' = e"',

    then its solution, obtained by integrating (4.14) three times, is (4.15) where c1, c2, c3 can take on any numerical values.

  • Lesson 4A MuLTIPLICITY OF SoLUTIONS OF A DIFFERENTIAL EQuATION 29

    Two conclusions seem to stem from these examples. First, if a differen-tial equation has a solution, it has infinitely many solutions (remember the c's can have infinitely many values). Second, if the differential equation is of the first order, its solution contains one arbitrary constant; if of the second order, its solution contains two arbitrary constants; if of the nth order, its solution contains n arbitrary constants. That both conjectures are in fact false can be seen from the following examples.

    E:rample 4.2. The first order differential equation

    (a) (y')2 + y2 = 0, also the second order differential equation

    (b)

    each has only the one solution y = 0. E:rample 4.21. The first order differential equation

    (a) IY'I + 1 = 0, also the second order differential equation

    (b) IY"I + 1 = 0, has no solution.

    E:rample 4.22. The first order differential equation

    (a) xy' = 1 has no solution if the interval I is -1 < x < 1. Formally one can solve (a) to obtain (b) y = log lxl + c, but this function is discontinuous at x = 0. By Definition 3.4, a solution must satisfy the differential equation for every x in I.

    Remark. If x < 0, then by (b) above (c) y = log ( -x) + c1, x < 0, is a valid solution of (a). And if x > 0, then by (b) (d) y = log x + c2, x > 0, is a valid solution of (a). The line x = 0, therefore, divides the plane into two regions; in one (c) is valid, in the other (d) is valid. There is no solution, however, if the region includes the line x = 0.

  • 30 BASIC CONCEPTS Chapter 1

    E:~tample 4.23. The first order differential equation

    (a) (y' - y)(y' - 2y) = 0 has the solution

    (b) which has two arbitrary constants instead of the usual one.

    These examples should warn you not to jump immediately to the con-clusion that every differential equation has a solution, or if it does have a solution that this solution w:ill contain arbitrary constants equal in num-ber to the order of the differential equation. It should comfort you to know, however, that there are large classes of differential equations for which the above conjectures are true, and that these classes include most of the equations which you are likely to encounter. For these classes only, then, we can assert: the solution of a differential equation of order n contains n arbitrary constants c 11 c2, , c,..

    It is customary to call a solution which contains n constants c11 c2, , c,. an n-parameter family of solutions, and to refer to the con-stants c1 to c,. as parameters. In this new notation, we would say (4.11) is a !-parameter family of solutions of (4.1); (4.13) is a 2-parameter family of solutions of (4.12), etc.

    Definition 4.3. The junctions defined by (4.31) y ,;,., f(x, ell c2, , c,.) of then+ 1 variables, x, c1, c2 , , c,. will be called an n-parameter family of solutions of the nth order differential equation

    (4.32) F(x,y,y', , y

  • Lesson 4B FINDING EQUATION FROM FAMILY oF SoLUTIONS 31

    second order differential equation

    (b) F(x,y,y',y") = y" - 3y' + 2y - 4x = 0. Solution. Let a,b be any two values of c1,c2 respectively. Then, by (a),

    (c) y = f(x) =; 2x + 3 + aez + be2"'. [Note that (c) now defines a function only of x.] The first and second derivatives of (c) are (d) y' = f'(x) = 2 + ae"' + 2be2z, y" = f"(x) = ae"' + 4be2"'. Substituting in (b) the values off, f' and f", as found in (c) and (d), for y, y', y", we obtain

    (e) F(x,f,f',f") = ae"' + 4be2"' - 6 - 3ae"' - 6be 2"' + 4x + 6 + 2ae"' + 2be2"' - 4x = 0.

    You can verify that the left side of (e) reduces to zero. Hence by Defini-tion 4.3, (a) is a 2-parameter family of solutions of (b).

    LESSON 4B. Method of Finding a Differential Equation if Its n-Parameter Family of Solutions Is Known. We shall now show you how to find the differential equation when its n-parameter family of solutions is known. You must bear in mind that although the family will contain the requisite number of n arbitrary constants, the nth order differential equation whose solution it is, contains no such constants. In solving problems of this type, therefore, these constants must be elimi-nated. Unfortunately a standard method of eliminating these constants is not always the easiest to use. There are frequently simpler methods which cannot be standardized and which will depend on your own in-genuity.

    Example 4.4. Find a differential equation whose 1-parameter family of solutions is

    (a) y = C COS X + X. Solution. In view of what we have already said, we assume that

    since (a) contains one constant, it is the solution of a first order differential equation. Differentiating (a), we obtain (b) y' = -c sin x + 1. This differential equation cannot be the one we seek since it contains the parameter c. To eliminate it, we multiply (a) by sin x, (b) by cos x and

  • 32 BASIC CONCEPTS

    add the equations. There results

    (c) y sin x + y' cos x = x sin x + cos x, (y' - 1) cos x + (y - x) sin x = 0,

    7r 37r y' = (x - y) tan x + 1, x ,&. 2 ' 2' '

    Chapter 1

    which is the required differential equation. [We could also have solved (b) for c and substituted its value in (a).] Note that the interval for which (a) is a solution of (c) must exclude certain points even though the function (a) is defined for these points.

    Example 4.5. Find a differential equation whose 2-parameter family of solutions is

    (a) Solution. Since (a) contains two parameters, we assume it is the

    solution of a second order differential equation. We therefore differentiate (a) twice, and obtain (b) (c)

    y' = Cte"' - c2e-"',

    y" = c1e"' + c2e-"'. Because of the presence of the constants c1 and c2 in (c), it cannot be the differential equation we seek. A number of choices are available for elimi-nating c1 and c2 We could, for example, solve (a) and (b) simultaneously for c1 and c2 and then substitute these values in (c). This method is a standard one which is always available to you, provided you know how to solve the pair of equations. An easier method is to observe that the right side of (c) is the same as the right side of (a). Hence, by equating their left sides, we have (d) y"- y = 0, which is the required differential equation.

    Example 4.51. Find a differential equation whose 2-parameter family of solutions is

    (a) Solution. Since (a) contains two constants, we assume it is the solu-

    tion of a second order differential equation. Hence we differentiate (a) twice, and obtain (b) (c)

    y' = c1 COS X - C2 sin X + 2x, y" = -c1 sin x - c2 cos x + 2

  • Lesson4C GENERAL AND PARTICULAR SoLUTIONs 33

    Here again you could use the standard method of finding c1 and c2 by solving (a) and (b) simultaneously, and then substituting these values in (c). [Or you could solve (b) and (c) simultaneously for c1 and c2 and substitute these values in (a)]. An easier method is to observe from (c) that (d) c1 sin x + c2 cos x = 2 - y" Substitution of (d) in (a) gives (e) y = 2 - y" + x2 or y" = x2 - y + 2, which is the required differential equation.

    E:rample 4.52. Find a differential equation whose 1-parameter family of solutions represents a family of circles with centers at the origin.

    Solution. Here the family of solutions is not given to us in the form of a mathematical equation. However, the family of circles with center at the origin is

    (a) Since (a) has only 1-parameter r, we assume. it is the solution of a first order differential equation. Hence we differentiate (a) once, and obtain (b) X+ yy' = 0, which is the required differential equation. Note that in this example the parameter r was eliminated in differentiating (a), and we were thus able to obtain the required differential equation immediately.

    LESSON 4C. General Solution. Particular Solution. Initial Con-ditions. An n-parameter family of solutions of an nth order differential equation has been called traditionally a "general" solution of the differen-tial equation. And the function which results when we give a definite set of values to the constants c11 c2, , Cn in the family has been called a "particular solution" of the differential equation.

    Traditionally then, for example, y = ce"' which is a 1-pars.meter family of solutions of y' - y = 0, would be called its general solution. And if we let c = -2, then y = -2e"' would be called a particular solution of the equation. It is evident that an infinite number of particular solutions can be obtained from a general solution: one for each value of c.

    A general solution, if it is to be worthy of its name, should contain aU solutions of the differential equation, i.e., it should be possible to obtain every particular solution by giving proper values to the constants c11 c2, , Cn. Unfortunately, there are differential equations which have solu-tions not obtainable from the n-parameter family no matter what values

  • 34 BAsic CoNCEPTS Chapter 1

    are given to the constants. For example, the first order differential equation

    (4.6) y = xy' + (y')2 has for a solution the !-parameter family

    (4.61) Traditionally, this solution, since it contains the required one parameter, would be called the general solution of (4.6). However, it is not the gen-eral solution in the real meaning of this term since it does not include every particular solution. The function

    (4.62) x2 y= --4 is also a solution of (4.6). (Verify it.) And you cannot obtain this func-tion from (4.61) no matter what value you assign to c. ((4.61) is a first degree equation; (4.62) is a second degree equation.]

    Unusual solutions of the type (4.62), i.e., those which cannot be ob-tained from ann-parameter family or the so-called general solution, have traditionally been called "singular solutions." We shall show below by examples that the use of these terms-general solution and singular solu-tion-in their traditional meanings is undesirable. Rather than being helpful in the study of differential equations, their use leads only to confusion.

    Consider for example the. first order differential equation

    (4.63) Its solution is

    1 (4.64) y = (x + c) 2 (Verify it.) But (4.63) has another solution (4.65) y = 0, which cannot be obtained from (4.64) by assigning any value to c. By the traditional definition, therefore, y = 0 would be called a singular solution of (4.63). However, we can also write the solution of (4.63) as

    c2. (4.651) y = (Cx + 1) 2 [Now verify that (4.651) is a solution of (4.63).] In this form, y = 0 is not a singular solution at all. It can be obtained from (4.651) by setting C = 0. Hence use of the traditional definitions for general solution and

  • Lesson 4C GENERAL AND PARTICULAR SoLUTIONs 35

    singular solution in this example leads us to the uncomfortable contradic-tion that a solution can be both singular and nonsingular, depending on the choice of representation of the 1-parameter family.

    Here is another example. The first order differential equation (4.652) (y' - y)(y' - 2y) = 0 has the following two distinct 1-parameter family of solutions (4.653) (4.654)

    y = c1e"',

    y = c2e2z.

    [Verify that each of these families satisfies (4.652).] If we call (4.653) the general solution of (4.652), as it should be called

    traditionally since it contains the required one parameter, then the entire family of functions (4.654) is, in the traditional sense, singular solutions. They cannot be obtained from (